Q.H2S is less acidic than H2Te. Why?

Group 16 elements (O, S, Se, Te, Po) share the outer configuration $ns^2np^4$, two electrons short of a stable octet, and their common oxidation states are $-2$, $+2$, $+4$ and $+6$, with the higher states becoming more stable for the heavier members as metallic character increases and electronegativity falls down the group (justifying the placement of O, S, Se, Te and Po together despite their differing hydride and oxide chemistry). First ionisation enthalpy in Group 16 is lower than in the corresponding Group 15 element of the same period because the extra electron in $np^4$ must pair up in an already singly-occupied orbital, and this electron-pairing repulsion makes it easier to remove than an electron from the extra-stable, exactly half-filled $np^3$ configuration of Group 15. Down the group, acidity of the hydrides $H_2E$ increases ($H_2O < H_2S < H_2Se < H_2Te$) as the H-E bond weakens with increasing atomic size, while thermal stability of these same hydrides decreases in the same order, since a longer, wea …